Question

The molar volume of $He$ at $10.1325$ MPa and $273$ K is $0.11075$ of its molar volume at $101.325$ kPa at $273$ K. Calculate the radius of helium atom. The gas is assumed to show real gas nature. Neglect the value of $a$ for $He$.

• A. $r=1.33A^o$
• B. $r=2.66A^o$
• C. $r=0.266A^o$
• D. $r=1.19A^o$

Answer 1

The correct option is A $r=1.33A^o$

Real gas equation for n=1,
$(P+\frac{a}{V^2})(V-b)=RT$

neglecting the value of 'a' according to the question,

$P(V-b)=RT$
Given, $P_1=10.1325MPa=100atm$, $T=273K$, $V_1=0.011075V_2$, $P_2=101.325kPa=1atm$

$P_1(V_1-b)=RT$
$100(0.011075V_2-b)=273R$.........(1)
Similarly,
$P_2(V_2-b)=RT$
$1(V_2-b)=273R$............(2)

From (1) and (2)
$1.1075V_2-100b=V_2-b$

$0.1075V_2=99b$
$V_2=921b$

Substitute the value of $V_2$ in (2),
$921b-b=273\times 0.0821$
$920b=22.4133$
$b=0.0243L-mo{l}^{-1}$

$b=4\times \text{Vol. occupied by 1 mole of the gas}$
$=4\times \text{no. of atoms of the gas in 1 mole}\times \frac{4}{3}\pi r^3$

$=4N_A\times \frac{4}{3}\pi r^3$ (Assuming shape of the atom spherical)
The radius of He atom is given by the following expression.

$r^3=\frac{3b}{16\pi N_A}=\frac{3\times 0.0243\times {10}^{-3}}{16\times 3.1416\times 6.023\times {10}^{23}}=2.41\times {10}^{-30}$ $[\because 1L={10}^{-3}{m}^3]$

$r=1.33\times {10}^{-10}m=1.33A^o$
Hence, option A is correct.